Answer:
the frequency of photons [tex]v = 1.509\times10^{39}Hz[/tex]
Explanation:
Given: first ionization energy of 1000 kJ/mol.
No. of moles of sulfur = 1 mole
[tex]\Delta E_1 = 1000KJ/mol[/tex]
We know that plank's constant
[tex]h = 6.626\times10^{-34} Js[/tex]
Let the frequency of photons be ν
Also we know that ΔE = hν
this implies ν = ΔE/h
[tex]= \frac{10^6J}{6.626\times10^{-34} Js}[/tex]
[tex]v = 1.509\times10^{39}Hz[/tex]
Hence, the frequency of photons [tex]v = 1.509\times10^{39}Hz[/tex]
